A Note on Bessel’s Inequality
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Dragomir, Sever S (2001) A Note on Bessel’s Inequality. RGMIA research report collection, 4 (2).
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S.S. DRAGOMIR
Abstract. A monotonicity property of Bessel’s inequality in inner product spaces is given.
1. Introduction
Let X be a linear space over the real or complex number fieldK. A mapping (·,·) :X×X →Kis said to be apositive hermitian formif the following conditions are satisfied:
(i) (αx+βy, z) =α(x, z) +β(y, z) for allx, y, z∈X andα, β∈K; (ii) (y, x) = (x, y) for all x, y∈X;
(iii) (x, x)≥0 for allx∈X.
If kxk := (x, x)12 , x∈ X denotes the semi-norm associated to this form and (ei)i∈I is an orthornormal family of vectors inX, i.e., (ei, ej) =δij (i, j∈I), then one has the following inequality [15]:
(1.1) kxk2≥X
i∈I
|(x, ei)|2 for allx∈X, which is well known in the literature as Bessel’s inequality.
Indeed, for every finite partH ofI, one has:
0 ≤
x−X
i∈H
(x, ei)ei
2
=
x−X
i∈H
(x, ei)ei, x−X
j∈H
(x, ej)ej
= kxk2−X
i∈H
|(x, ei)|2−X
j∈H
|(x, ej)|2+ X
i,j∈H
(x, ei) (ej, x)δij
= kxk2−X
i∈H
|(x, ei)|2, for allx∈X,which proves the assertion.
The main aim of this paper is to improve this result as follows.
2. Results The following theorem holds.
Theorem 1. Let X be a linear space and (·,·)2,(·,·)1 two hermitian forms on X such that k·k2 is greater than or equal to k·k1, i.e., kxk2 ≥ kxk1 for all x ∈ X. Assume that(ei)i∈I is an orthornormal family in(X; (·,·)2)and(fi)i∈J is an
Date: January 05, 2000.
1991Mathematics Subject Classification. Primary 26D15; Secondary 46C99.
Key words and phrases. Bessel’s inequality, Inner product spaces.
1
2 S.S. DRAGOMIR
orthornormal family in(X; (·,·)1)such that for anyi∈Ithere exists a finiteK⊂J so that
(F) ei =X
j∈K
αjfj, αj ∈K (j∈K), then one has the inequality:
(2.1) kxk22−X
i∈I
|(x, ei)2|2≥ kxk21−X
j∈J
(x, fj)1
2≥0,
for allx∈X.
In order to prove this fact, we require the following lemma.
Lemma 1. Let X be a linear space endowed with a positive hermitian form (·,·) and(gk)k=1,n be an orthornormal family in(X; (·,·)). Then
(2.2)
x−
n
X
k=1
λkgk
2
≥ kxk2−
n
X
k=1
|(x, gk)|2≥0, for allλk∈Kandx∈X (k= 1, ..., n).
Proof. We will prove this fact by induction over “n”.
Supposen= 1. Then we must prove that
kx−λ1g1k2≥ kxk2− |(x, g1)|2, x∈X, λ1∈K.
A simple computation shows that the above inequality is equivalent with
|λ1|2−2 Re (x, λ1g1) +|(x, g1)|2≥0, x∈X, λ1∈K. Since Re (x, λ1g1)≤ |(x, λ1g1)|, one has
|λ1|2−2 Re (x, λ1g1) +|(x, g1)|2 ≥ |λ1|2−2|λ1| |(x, g1)|+|(x, g1)|2
≥ (|λ1| − |(x, g1)|)2≥0 for allλ1∈Kandx∈X, which proves the statement.
Now, assume that (2.2) is valid for “(n−1)”. Then we have:
x−
n
X
k=1
λkgk
2
=
(x−λngn)−
n−1
X
k=1
λkgk
≥ kx−λngnk2−
n−1
X
k=1
|(x−λngn, gk)|2
= kx−λngnk2−
n−1
X
k=1
|(x, gk)|2≥ kxk2− |(x, gn)|2−
n−1
X
k=1
|(x, gk)|2
= kxk2−
n
X
k=1
|(x, gk)|2,
for allλk ∈K,x∈X (k= 1, ..., n),and the proof of the lemma is complete.
Proof. (Theorem) LetH be a finite part of I. Sincek·k2 is greater than k·k1, we have:
kxk22−X
i∈H
|(x, ei)2|2 =
x−X
i∈H
(x, ei)2ei
2
2
≥
x−X
i∈H
(x, ei)2ei
2
1
, x∈X.
Since, by (F), we may state that for anyi∈H there exists a finiteK⊂J with ei= X
j∈K
(ei, fj)1fj, we have
x−X
i∈H
(x, ei)2ei
2
1
=
x−X
i∈H
(x, ei)2X
j∈K
(ei, fj)1fj
2
1
=
x−X
j∈K
X
i∈H
(x, ei)2ei, fj
!
1
fj
2
1
for allx∈X.
Applying the above lemma for (·,·) = (·,·)1, (gk)k=1,n = (fj)j∈K, we can conclude that
x−X
j∈K
λjfj
2
1
≥ kxk21−X
j∈K
(x, fj)1
2, x∈X,
where
λj = X
i∈H
(x, ei)2ei, fj
!
1
∈K (j∈K). Consequently, we have:
kxk22−X
i∈H
|(x, ei)2|2≥ kxk21−X
j∈K
(x, fj)1
2≥ kxk21−X
j∈J
(x, fj)1
2
for allx∈X andH a finite part ofI, from where results (2.1).
The proof is thus completed.
Corollary 1. Let k·k1,k·k2:X →R+be as above. Then for allx, y∈X, we have the inequality:
(2.3) kxk22kyk22− |(x, y)2|2≥ kxk21kyk21− |(x, y)1|2≥0, which is an improvement of the well known Cauchy-Scwartz inequality.
Proof. Ifkyk2= 0,then (2.3) holds with equality.
If kyki 6= 0, (i= 1,2), then for {e1} = n y
kyk2
o
, {f1} = n y
kyk1
o
, the above theorem yields that
kxk22kyk22− |(x, y)2|2
kyk22 ≥ kxk21kyk21− |(x, y)1|2 kyk21
4 S.S. DRAGOMIR
and sincekyk2≥ kyk1,the inequality (2.3) is obtained.
Remark 1. For a different proof of (2.3), see also [5].
Now, we will give some natural applications of the above theorem.
3. Applications
(1) Let (X; (·,·)) be an inner product space and (ei)i∈I an orthornormal family in X. Assume that A:X →X is a linear operator such thatkAxk ≤ kxk for all x ∈ X and (Aei, Aej) = δij for all i, j ∈ I. Then one has the inequality
kxk2−X
i∈I
|(x, ei)|2≥ kAxk2−X
i∈I
|(Ax, Aei)|2≥0 for allx∈X.
The proof follows by the hermitian forms (x, y)2 = (x, y) and (x, y)1 = (Ax, Ay) forx, y∈X and for the family (fi)i∈I = (ei)i∈I.
(2) If A : X → X is such that kAxk ≥ kxk for all x ∈ X, then, with the previous assumptions, we also have
0≤ kxk2−X
i∈I
|(x, ei)|2≤ kAxk2−X
i∈I
|(Ax, Aei)|2, for allx∈X.
(3) Suppose that A : X → X is a symmetric positive definite operator with (Ax, x) ≥ kxk2 for all x∈ X. If (ei)i∈I is an orthornormal family in X such that (Aei, Aej) =δij for alli, j∈I, then one has the inequality
0≤ kxk2−X
i∈I
|(x, ei)|2≤(Ax, x)−X
i∈I
|(Ax, ei)|2, for allx∈X.
The proof follows from the above theorem for the choices (x, y)1= (Ax, y) and (x, y)2= (x, y), x, y∈X. We omit the details.
For other inequalities in inner product spaces, see the papers [1]-[14] and [7]-[6]
where further references are given.
References
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[3] S.S. DRAGOMIR and B. MOND, On the Boas-Bellman generalisation of Bessel’s inequality in inner product spaces,Italian J. of Pure and Appl. Math.,3(1998), 29-38.
[4] S.S. DRAGOMIR and B. MOND, On the superadditivity and monotonicity of Gram’s in- equality and related results,Acta Math. Hungarica,71(1-2) (1996), 75-90.
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